question_answer

A particle of mass 2 kg is on a smooth horizontal table and moves in a circular path of radius 0.6 m. The height of the table from the ground is 0.8 m. If the angular speed of the particle is 12 rad \[{{s}^{-1}}\], the magnitude of its angular momentum about a point on the ground right under the centre of the circle is

A) \[14.4\,kg\,{{m}^{2}}{{s}^{-1}}\]

B) \[8.64\,kg\,{{m}^{2}}{{s}^{-1}}\]

C) \[20.16\,kg\,{{m}^{2}}{{s}^{-1}}\]

D) \[11.52\,kg\,{{m}^{2}}{{s}^{-1}}\]

Correct Answer: A

Solution :

[a] Angular momentum, \[{{L}_{0}}=mvr\,\sin \,{{90}^{o}}\] \[=2\times 0.6\times 12\times 1\times 1\] [\[As\,V=r\omega ,\,Sin\,{{90}^{o}}=1\]]   \[So,\,\,{{L}_{0}}=14.4\,kg{{m}^{2}}/s\]